Question

the grade point average collected from a random sample of 150 students. assume that the population standard deviation is 0.78. find the margin of error if c = 0.98.

a. 0.11
b. 0.12
c. 0.08
d. 0.15

Explanation:

Step1: Find the z - value

For a confidence level \(c = 0.98\), the significance level \(\alpha=1 - c=1 - 0.98 = 0.02\). Then \(\frac{\alpha}{2}=\frac{0.02}{2}=0.01\).
The \(z\) - value \(z_{\frac{\alpha}{2}}\) is the \(z\) - score such that the area to the right of \(z_{\frac{\alpha}{2}}\) is \(0.01\). Using the standard normal distribution table or a calculator, \(z_{\frac{\alpha}{2}}=z_{0.01}\approx 2.33\).

Step2: Calculate the margin of error formula

The formula for the margin of error \(E\) when the population standard deviation \(\sigma\) is known is \(E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\).
Given \(\sigma = 0.78\), \(n = 150\), and \(z_{\frac{\alpha}{2}}=2.33\).
Substitute the values into the formula: \(E=2.33\times\frac{0.78}{\sqrt{150}}\).
First, calculate \(\sqrt{150}\approx12.25\). Then \(\frac{0.78}{12.25}\approx0.0637\).
Finally, \(E = 2.33\times0.0637\approx0.15\).

Answer:

D. \(0.15\)